Question 980796: I'm having difficulties solving this problem. Could someone please help me?
Two forces with magnitudes of 131 and 52 newtons act on point with a resultant force of 100
newtons. Find the angle, in radians, between the forces.
Enter answer accurate to four decimal places.
Thank you.
Found 2 solutions by Timnewman, KMST:
Answer by Timnewman(323)   (Show Source):
You can put this solution on YOUR website! From your cosine rule,recall that
CosB=(c²+a²-b²)/2ac
Draw the vector and apply the formula.
from your diagramme,(i.e triangle) formed by the two vectors,
if a=52,
b=100
C=131
B=?
then,
cosB=(131²+52²-100²)/2*131*52
evaluating the above,
cosB=(9865)/(13624)
cosB=0.7241
B=cos-1(0.7241)
B=43.6068degres
now convert 43.6068º to radian
B=(pi/180)*46.6068
B=0.7612radian
the angle between them is 0.7612rad
HOPE THIS HELPS?
Answer by KMST(5328)  (Show Source):
You can put this solution on YOUR website! While I was busy drawing pretty pictures, Tim was posting a more succinct answer.
I am posting my answer too, hoping that my drawings, and my different angle (pun intended) help too.
 I am assuming that you have to find the angle  ,
but we can also find the other angle in the parallelogram,
which is supplementary to :  .
The resultant divides that parallelogram into two (congruent) triangles.
We need to apply to one of those triangles the law of cosines,
which is sort-of like an extension of the Pythagorean theorem,
for triangles that are not necessarily right triangles.
For  law of cosines says  .
Applying law of cosines to the triangle above, with  , we get
 and form that we get
 (rounded)
Since they are supplementary angles  (rounded) .
Inverse cosine function tells us that the angles, in radians are
 and 
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